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Packing fraction of body-centered cubic lattice - solid state physics

by cameo_demon
Tags: crystal, lattice, solid state
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Sep10-08, 02:41 PM
P: 15
1. The problem statement, all variables and given/known data

This is part of a series of short questions (i.e. prove everything in Kittel Ch. 1, Table 2):

Prove that the packing fraction of a BCC (body-centered) cubic lattice is:

1/8 * pi * \sqrt{3}

2. Relevant equations

packing fraction = volume of a sphere / volume of primitive cell

3. The attempt at a solution

each lattice point (there are two total for BCC) can hold a sphere (or at least part of one) with radius a\sqrt{3} / 2. subbing in:

2 * 4/3 * pi * (a\sqrt{3} / 2) ^{3} / (a^{3}/2)

the \sqrt{3} becomes a 9. how does the \sqrt{3} possibly remain?

my question: how the heck did kittel get 1/8 * pi * \sqrt{3}?
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Sep29-08, 10:08 PM
P: 1
This does not solve your problem at all, but at least clarify how the \sqrt3 survives.
(\sqrt3)^3 = (\sqrt3)^2 * \sqrt3 = 3* \sqrt3.
Nevertheless, you can not get the answer by means of your assupmtion. The packing fraction is the maximum proportion of the available volume in the cell that can be filled with spheres... keep it in mind.

Indeed, the radius of such sphere is half of the nearest-neighbor distance. Try it!
Steve Dutch
Feb20-10, 08:38 PM
P: 1
It's basically geometry. Let the unit cell be a cube of side 1. The long diagonal has length sqrt(3). The corner atoms and the central atoms all have their centers on the long diagonal and touch, so their radius is sqrt(3)/4. So what's the volume of an atom, and how many atoms are in a unit cell (remember the corner atoms are shared with neighboring cells)?

Jun2-11, 01:49 AM
P: 5
Packing fraction of body-centered cubic lattice - solid state physics

Interesting. What happens if the spheres are allowed to have two different sizes? Would it be possible to exceed the FCC density (0.74)?

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